Answer
$\dfrac{1995}{64}$
Work Step by Step
since,we have $\sum_{i=1}^{6}(\dfrac{3}{2})^i$
Here, terms are: {$\dfrac{3}{2},\dfrac{9}{4},\dfrac{27}{8}...$}; which is an geometric sequence.
Their sum is $S_n=\dfrac{a_1(1-r^n)}{1-r}$
Thus,
$S_6=\dfrac{(\dfrac{3}{2})(1-(\dfrac{3}{2})^6)}{1-\dfrac{3}{2}}$
or, $=\dfrac{(\dfrac{3}{2})(\dfrac{-665}{64})}{-\dfrac{1}{2}}$
Hence,
$=\dfrac{1995}{64}$
